Straight Lines (Circumcenter) Problems
1) Find the circumcenter of the triangle whose vertices are (1, 0), (-1, 2) and (3, 2).
Solution: Vertices of the triangle are A (1, 0), B (-1, 2), C (3, 2).
Let S (x, y) be the circumcenter of triangle ABC,
Then SA = SB = SC,
Let SA = SB ⇒ SA² = SB²,
(x – 1)² + y² = (x + 1)²+ (y – 2)²,
⇒ x² – 2y + 1 + y² = x² + 2x + 1 + y² – 4y + 4.
⇒ 4x – 4y = -4.
⇒ x – y = -1 … (1)
SB = SC ⇒ SB² = SC²
(x + 1)² + (y – 2)² = (x – 3)² + (y – 2)²
⇒ x² + 2x + 1 + y² – 4y + 4 = x² – 6x + 9 + y² – 4y + 4
2x + 1 = -6x + 9
8x = 8 ⇒ x = 1
From equation (1)
x – y = – 1
1 – y = -1
y = 2,
Therefore circumcenter (x, y) = (1, 2).
2) Find the circumcenter of the triangle formed by the straight lines x + y = 0, 2x + y + 5 = 0 and x – y = 2.
Solution: Let the equation of AB be x + y = 0 … (1)
BC be 2x + y + 5 = 0 … (2)
And AC be x – y = 2 … (3)
Solving (1) and (2), vertex B = (-5, 5)
Solving (2) and (3), vertex C = (-1, -3)
Solving (1) and (3), vertex A = (1, -1)
Let S (x, y) be the circumcenter of triangle ABC
Then SA = SB = SC
SA = SB
SA² = SB²
(x + 5)² + (y – 5)² = (x + 1)² + (y + 3)²
8x – 16y = -40
x – 2y = -5 … (4)
SA = SC
SA² = SC²
(x + 1)² + (y + 3)² = (x – 1)² + (y + 1)²
4x + 4y = -8
x + y = -2 … (6)
solving (4) and (5) point of intersection is (-3, 1)
circumcenter is S (-3, 1).